Question

Find the slope of tangent and normal to the curve $x^2+2y+y^2=0$ at $(x_1,y_1)$

Answer 1

Given Curve$x^2+2y+y^2=0$

$\Rightarrow 2x+2\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$\Rightarrow x+\frac{dy}{dx}+y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}+y\frac{dy}{dx}=-x$

$\Rightarrow (1+y)\frac{dy}{dx}=-x$

$\Rightarrow \frac{dy}{dx}=\frac{-x}{(1+y)}$

Slope of tangent at $(x_1,y_1)=\frac{-x_1}{(1+y_1)}$

Slope of normal$=\frac{-1}{\frac{dy}{dx}}$ at $(x_1,y_1)=\frac{-1}{\frac{-x_1}{(1+y_1)}}=\frac{1+y_1}{x_1}$